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Prove by Mathematical Induction 14| 3^(4n+2) +5^(2n+1)

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ericmilamattim | (Level 1) Honors

Posted September 7, 2013 at 11:53 AM via web

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Prove by Mathematical Induction

14| 3^(4n+2) +5^(2n+1)

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted September 8, 2013 at 4:02 AM (Answer #2)

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Prove that `14|(3^(4n+2)+5^(2n+1))` :

Base step: This is true for n=0 since 14|(9+5); it is also true for n=1 since 14|(729+125)

Assume that for some `k in NN` that `14|(3^(4k+2)+5^(2k+1))`

Show that `14|(3^(4(k+1)+2)+5^(2(k+1)+1))` or `14|(3^(4k+6)+5^(2k+3))` :

`3^(4k+6)+5^(2k+3)=81*3^(4k+2)+25*5^(2k+3)`

Now `14|(3^(4k+2)+5^(2k+1))==> 3^(4k+2)+5^(2k+1)-=0("mod"14)`

Then `(3^(4k+2)("mod"14)+5^(2k+1)("mod"14))("mod"14)-=0("mod"14)` ``So `3^(4k+2)("mod"14)-=-5^(2k+1)("mod"14)`

`(81*3^(4k+2)("mod"14)+25*5^(2k+1)("mod"14))("mod"14)`

`-=(81(-5^(2k+1))("mod"14)+25*5^(2k+1)("mod"14))("mod"14)`

`-=((-81+25)(5^(2k+1)))("mod"14)`

`-=((-56)(5^(2k+1))("mod"14)`

`-=(-56("mod"14)*(5^(2k+1))("mod"14))("mod"14)`

`-=0("mod"14)`

`==>14|(81*3^(4k+2)+25*5^(2k+1))` as required.

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